Java中的HashMap和Map对象之间有什么区别？

Question

What is the difference between the following maps I create (in another question, people answered using them seemingly interchangeably and I'm wondering if/how they are different):

HashMap<String, Object> map = new HashMap<String, Object>();
Map<String, Object> map = new HashMap<String, Object>();

Answer 1

There is no difference between the objects; you have a HashMap<String, Object> in both cases. There is a difference in the interface you have to the object. In the first case, the interface is HashMap<String, Object> , whereas in the second it's Map<String, Object> . But the underlying object is the same.

The advantage to using Map<String, Object> is that you can change the underlying object to be a different kind of map without breaking your contract with any code that's using it. If you declare it as HashMap<String, Object> , you have to change your contract if you want to change the underlying implementation.

---

Example: Let's say I write this class:

class Foo {
    private HashMap<String, Object> things;
    private HashMap<String, Object> moreThings;

    protected HashMap<String, Object> getThings() {
        return this.things;
    }

    protected HashMap<String, Object> getMoreThings() {
        return this.moreThings;
    }

    public Foo() {
        this.things = new HashMap<String, Object>();
        this.moreThings = new HashMap<String, Object>();
    }

    // ...more...
}

The class has a couple of internal maps of string->object which it shares (via accessor methods) with subclasses. Let's say I write it with HashMap s to start with because I think that's the appropriate structure to use when writing the class.

Later, Mary writes code subclassing it. She has something she needs to do with both things and moreThings , so naturally she puts that in a common method, and she uses the same type I used on getThings / getMoreThings when defining her method:

class SpecialFoo extends Foo {
    private void doSomething(HashMap<String, Object> t) {
        // ...
    }

    public void whatever() {
        this.doSomething(this.getThings());
        this.doSomething(this.getMoreThings());
    }

    // ...more...
}

Later, I decide that actually, it's better if I use TreeMap instead of HashMap in Foo . I update Foo , changing HashMap to TreeMap . Now, SpecialFoo doesn't compile anymore, because I've broken the contract: Foo used to say it provided HashMap s, but now it's providing TreeMaps instead. So we have to fix SpecialFoo now (and this kind of thing can ripple through a codebase).

Unless I had a really good reason for sharing that my implementation was using a HashMap (and that does happen), what I should have done was declare getThings and getMoreThings as just returning Map<String, Object> without being any more specific than that. In fact, barring a good reason to do something else, even within Foo I should probably declare things and moreThings as Map , not HashMap / TreeMap :

class Foo {
    private Map<String, Object> things;             // <== Changed
    private Map<String, Object> moreThings;         // <== Changed

    protected Map<String, Object> getThings() {     // <== Changed
        return this.things;
    }

    protected Map<String, Object> getMoreThings() { // <== Changed
        return this.moreThings;
    }

    public Foo() {
        this.things = new HashMap<String, Object>();
        this.moreThings = new HashMap<String, Object>();
    }

    // ...more...
}

Note how I'm now using Map<String, Object> everywhere I can, only being specific when I create the actual objects.

If I had done that, then Mary would have done this:

class SpecialFoo extends Foo {
    private void doSomething(Map<String, Object> t) { // <== Changed
        // ...
    }

    public void whatever() {
        this.doSomething(this.getThings());
        this.doSomething(this.getMoreThings());
    }
}

...and changing Foo wouldn't have made SpecialFoo stop compiling.

Interfaces (and base classes) let us reveal only as much as is necessary , keeping our flexibility under the covers to make changes as appropriate. In general, we want to have our references be as basic as possible. If we don't need to know it's a HashMap , just call it a Map .

This isn't a blind rule, but in general, coding to the most general interface is going to be less brittle than coding to something more specific. If I'd remembered that, I wouldn't have created a Foo that set Mary up for failure with SpecialFoo . If Mary had remembered that, then even though I messed up Foo , she would have declared her private method with Map instead of HashMap and my changing Foo 's contract wouldn't have impacted her code.

Sometimes you can't do that, sometimes you have to be specific. But unless you have a reason to be, err toward the least-specific interface.

Answer 2

Map is an interface that HashMap implements. The difference is that in the second implementation your reference to the HashMap will only allow the use of functions defined in the Map interface, while the first will allow the use of any public functions in HashMap (which includes the Map interface).

It will probably make more sense if you read Sun's interface tutorial

Answer 3

Map has the following implementations:

1. HashMap Map m = new HashMap();

2. LinkedHashMap Map m = new LinkedHashMap();

3. Tree Map Map m = new TreeMap();

4. WeakHashMap Map m = new WeakHashMap();

Suppose you have created one method (this is just pseudocode).

public void HashMap getMap(){
   return map;
}

Suppose your project requirements change:

1. The method should return map contents - Need to return HashMap .
2. The method should return map key's in insertion order - Need to change return type HashMap to LinkedHashMap .
3. The method should return map key's in sorted order - Need to change return type LinkedHashMap to TreeMap .

If your method returns specific classes instead of something that implements the Map interface, you have to change the return type of getMap() method each time.

But if you use the polymorphism feature of Java, and instead of returning specific classes, use the interface Map , it improves code reusability and reduces the impact of requirement changes.

Answer 4

I was just going to do this as a comment on the accepted answer but it got too funky (I hate not having line breaks)

ah, so the difference is that in general, Map has certain methods associated with it. but there are different ways or creating a map, such as a HashMap, and these different ways provide unique methods that not all maps have.

Exactly--and you always want to use the most general interface you possibly can. Consider ArrayList vs LinkedList. Huge difference in how you use them, but if you use "List" you can switch between them readily.

In fact, you can replace the right-hand side of the initializer with a more dynamic statement. how about something like this:

List collection;
if(keepSorted)
    collection=new LinkedList();
else
    collection=new ArrayList();

This way if you are going to fill in the collection with an insertion sort, you would use a linked list (an insertion sort into an array list is criminal.) But if you don't need to keep it sorted and are just appending, you use an ArrayList (More efficient for other operations).

This is a pretty big stretch here because collections aren't the best example, but in OO design one of the most important concepts is using the interface facade to access different objects with the exact same code.

Edit responding to comment:

As for your map comment below, Yes using the "Map" interface restricts you to only those methods unless you cast the collection back from Map to HashMap (which COMPLETELY defeats the purpose).

Often what you will do is create an object and fill it in using it's specific type (HashMap), in some kind of "create" or "initialize" method, but that method will return a "Map" that doesn't need to be manipulated as a HashMap any more.

If you ever have to cast by the way, you are probably using the wrong interface or your code isn't structured well enough. Note that it is acceptable to have one section of your code treat it as a "HashMap" while the other treats it as a "Map", but this should flow "down". so that you are never casting.

Also notice the semi-neat aspect of roles indicated by interfaces. A LinkedList makes a good stack or queue, an ArrayList makes a good stack but a horrific queue (again, a remove would cause a shift of the entire list) so LinkedList implements the Queue interface, ArrayList does not.

Answer 5

As noted by TJ Crowder and Adamski, one reference is to an interface, the other to a specific implementation of the interface. According to Joshua Block, you should always attempt to code to interfaces, to allow you to better handle changes to underlying implementation - ie if HashMap suddenly was not ideal for your solution and you needed to change the map implementation, you could still use the Map interface, and change the instantiation type.

Answer 6

In your second example the "map" reference is of type Map , which is an interface implemented by HashMap (and other types of Map ). This interface is a contract saying that the object maps keys to values and supports various operations (eg put , get ). It says nothing about the implementation of the Map (in this case a HashMap ).

The second approach is generally preferred as you typically wouldn't want to expose the specific map implementation to methods using the Map or via an API definition.

Answer 7

Map is the static type of map, while HashMap is the dynamic type of map. This means that the compiler will treat your map object as being one of type Map, even though at runtime, it may point to any subtype of it.

This practice of programming against interfaces instead of implementations has the added benefit of remaining flexible: You can for instance replace the dynamic type of map at runtime, as long as it is a subtype of Map (eg LinkedHashMap), and change the map's behavior on the fly.

A good rule of thumb is to remain as abstract as possible on the API level: If for instance a method you are programming must work on maps, then it's sufficient to declare a parameter as Map instead of the stricter (because less abstract) HashMap type. That way, the consumer of your API can be flexible about what kind of Map implementation they want to pass to your method.

Answer 8

You create the same maps.

But you can fill the difference when you will use it. With first case you'll be able to use special HashMap methods (but I don't remember anyone realy useful), and you'll be able to pass it as a HashMap parameter:

public void foo (HashMap<String, Object) { ... }

...

HashMap<String, Object> m1 = ...;
Map<String, Object> m2 = ...;

foo (m1);
foo ((HashMap<String, Object>)m2); 

Answer 9

Map是接口，Hashmap是实现Map Interface的类

Answer 10

Adding to the top voted answer and many ones above stressing the "more generic, better", I would like to dig a little bit more.

Map is the structure contract while HashMap is an implementation providing its own methods to deal with different real problems: how to calculate index, what is the capacity and how to increment it, how to insert, how to keep the index unique, etc.

Let's look into the source code:

In Map we have the method of containsKey(Object key) :

boolean containsKey(Object key);

JavaDoc:

boolean java.util.Map.containsValue(Object value)

Returns true if this map maps one or more keys to the specified value. More formally, returns true if and only if this map contains at least one mapping to a value v such that (value==null ? v==null : value.equals(v)) . This operation will probably require time linear in the map size for most implementations of the Map interface.

Parameters:value

value whose presence in this map is to betested

Returns:true

if this map maps one or more keys to the specified

valueThrows:

ClassCastException - if the value is of an inappropriate type for this map (optional)

NullPointerException - if the specified value is null and this map does not permit null values (optional)

It requires its implementations to implement it, but the "how to" is at its freedom, only to ensure it returns correct.

In HashMap :

public boolean containsKey(Object key) {
    return getNode(hash(key), key) != null;
}

It turns out that HashMap uses hashcode to test if this map contains the key. So it has the benefit of hash algorithm.

Answer 11

Map is the Interface and Hashmap is the class that implements that.

So in this implementation you create the same objects

Answer 12

HashMap是Map的实现，因此完全相同，但是具有“ clone（）”方法，如我在参考指南中看到的那样））

Answer 13

HashMap<String, Object> map1 = new HashMap<String, Object>();
Map<String, Object> map2 = new HashMap<String, Object>();  

First of all Map is an interface it has different implementation like - HashMap , TreeHashMap , LinkedHashMap etc. Interface works like a super class for the implementing class. So according to OOP's rule any concrete class that implements Map is a Map also. That means we can assign/put any HashMap type variable to a Map type variable without any type of casting.

In this case we can assign map1 to map2 without any casting or any losing of data -

map2 = map1